Proposition 8

Let $X$ be a metrizable space and let $(A_n)$ be a sequence of Souslin subspaces of $X$. Then the union and the intersection of the $A_n$ are Souslin spaces.

These subspaces are certainly metrizable. The existence of the canonical map of the sum of the $A_n$ onto the subspace $\bigcup_n A_n$ of $X$ shows that the latter is a Souslin space; and $\bigcap_n A_n$ is Souslin by virtue of Propositions 5 and 7 and Lemma 1 of no. 1.

In general, even in a Polish space, the complement of a Souslin subspace is not necessarily Souslin (cf. Exercise 6); see, however, no. 6, Theorem 2, Corollary.